0
votes
1answer
19 views

The lattice of closed subsets of an algebraic closure operator is an algebraic lattice

Let $A$ be a set. Let $C:Su(A)\longrightarrow Su(A)$ be a function, where $Su(A)$ denotes the set of all subsets of $A$. Suppose that 1) $X\subseteq C(X)$ 2) $X\subseteq Y\rightarrow C(X)\subseteq ...
1
vote
1answer
37 views

Distributive lattices and Birkhoff theorem

I am trying to prove the teorem (Birkhoff) $L$ is a nondistributive lattice iff $M_5$ or $N_5$ can be embedded into $L$ The only part of the proof which I can't understand is this (I am copying from ...
3
votes
2answers
84 views

Examples of Stone algebras which are not Boolean algebras

Grätzer, in his Lattice Theory: Foundation, describes a Stone algebra as a distributive lattice with pseudocomplementation $L$ which satisfies the Stone identity: for every $a \in L$, $\neg a \vee ...
1
vote
2answers
45 views

When do DeMorgan's laws hold in a Heyting algebra

I'm working a bit with Heyting algebras (which are pseudocomplemented distributive lattives, right?) and I have a question about DeMorgan's laws. I know that, in general, it's not the case that $-(X ...
0
votes
0answers
27 views

Congruence lattice of N5

I calculated the the congruence lattice of $N_5$ using hit and trial and then verified it with Universal Algebra calculator. But I need to prove that it is the congruence lattice of $N_5$ How should I ...
3
votes
1answer
103 views

Semilattices are congruence-semi-distributive

A semilattice $(S,\cdot)$ is a commutative idempotent semigroup. A congruence on a semilattice is an equivalence relation that preserves multiplication, i.e. $x_1\mathrel{\theta} y_1$ and ...
1
vote
1answer
51 views

Factor congruences of non trivial Lattices

A pair of congruences $\theta$ and $\theta^*$ are called factor congruences if $\theta \vee \theta^*$ = full congruence. $\nabla$ $\theta \wedge \theta^*$ = trivial congruence. $\triangle$ I need ...
0
votes
0answers
17 views

A filterbase generating filter F

Show that a non empty subset $X$ of a filter $F$ in $B$ is a base for $F$ iff $X$ generates $F$ and for all $x,y$ $\in$ $X$ $\exists$ $z $ $\in$ $X$ such that $z$ $\leqq$ x $\wedge$ y.
0
votes
2answers
36 views

How is modularity a weakened form of distributivity?

While reading an essay Lattice Theory- Its Birth and Life, the following line confused me: modularity is a weakened form of distributivity Just to be clear, here modularity and distributivity of ...
3
votes
1answer
38 views

Confirmation needed of the fact that subcategory $\mathbf{Lat}$ is not full in $\mathbf{Pos}$

If you are familiar with this stuff then you probably don't need the information I have added. So let me start with the question: Can you prove that category $\mathbf{Lat}$ is not a full ...
5
votes
1answer
110 views

$\mathbf{N}_5$ as a congruence lattice

A finite lattice is said to be representable if there exists a finite algebra whose congruence lattice is isomorphic to that lattice. As I was reading a paper, I came across the line: "The reader can ...
4
votes
1answer
73 views

The function $f(x)=(x\vee a)\wedge b$ in a lattice.

Is there an algebraic modular lattice $(X,\vee,\wedge)$ and $a,b\in X$ with $a\le b$ such that the function $$f:X\to X$$ $$f(x)=(x\vee a)\wedge b$$ is not $\vee$-homomorphism?
1
vote
0answers
69 views

Congruence lattice of a partial algebra is algebraic

A partial operation on a nonempty set $A$ is a map $f:\mathrm{dom}(f,A)\to A$ where $\mathrm{dom}(f,A)\subseteq A^n$ for some $n\in\mathbb{N}$. A partial algebra is an ordered pair $(A,P)$ where $A$ ...
1
vote
2answers
54 views

Notation for “incommensurate” elements?

Say that $x\wedge y\not=x,y$, that is neither $x\leq y$ nor $y\leq x$. Is there a way to denote this? I've been saying $x<>y$ but that's completely made up.
0
votes
1answer
217 views

What is a subdirect product?

I'm having trouble understanding what a subdirect product is. Say $G$ is a subdirect product of $H=\prod H_i$ - this means that the homomorphisms $f_i:G\to H_i$ are surjective, which can be ...
1
vote
2answers
106 views

The kernel of the kernel.

From Wikipedia-Entry on Equivalence Relatin:Lattices The possible equivalence relations on any set X, when ordered by set inclusion, form a complete lattice, called Con X by convention. The ...
2
votes
2answers
173 views

Smallest Congruence Relation generated by a set

$\newcommand{\cl}{\operatorname{cl}}$ Let $R \subset S \times S$ be a binary relation, the smallest i) reflexive relation containing it is $$ \cl_\mathrm{ref} = R \cup \{ (x,x) : x \in S \} $$ ii) ...
1
vote
1answer
62 views

Proof that the lattice of fully invariant congruences is a sublattice of the lattice of all congruences

Let $\mathfrak U$ be an algebra (i.e. a set, called universe, together with several $n$-ary operations) in the sense of universal algebra. Denote by $\operatorname{Con} \mathfrak U$ the set of all ...
0
votes
1answer
84 views

Represent the three element chain as a subdirect product of subdirectly irreducible lattices.

Represent the three element chain as a subdirect product of subdirectly irreducible lattices. I know the two element chain as either a Boolean algebra or a semilattice,is subdirectly irreducible.In ...
0
votes
0answers
35 views

Question about a property of lattice-morphism

I would like to know if there is a name for the class of commutative (i.e., $\phi(x,y)=\phi(y,x)$) lattice-morphisms $\phi : L_1\times L_{1} \rightarrow L_2$ with the following property: $\phi(x ...
4
votes
2answers
161 views

Stone duality for ideals and filters (exercise)

In A Course in Universal Algebra (Burris, Sankapannavar), the exercise 4.4.7-8, p.158, says: Let $A$ be a Boolean algebra. Denote $A^\ast:=\{\text{ultrafilters of }A\}$, and give $A^\ast$ the ...
2
votes
1answer
148 views

correspondence for universal subalgebras of $U/\vartheta$

Let $U$ be a universal algebra of type $T$, and denote $\mathrm{Con}(U)\!=\!\{\text{congruence relations on }U\}$ and $\mathrm{Sub}(U)\!=\!\{\text{subalgebras of }U\}$. Let "$\leq$" mean "subalgebra". ...
3
votes
1answer
130 views

$M_3$ is a simple lattice

I'd like to prove (exercise 9.5 in Roman's Lattices and Ordered Sets, p.203) that the lattice $M_3$ is simple, meaning that the only congruences on $M_3$ are the trivial ones (the 'equality' ...
2
votes
1answer
112 views

ideals of a ring form a modular lattice

We know that if $M$ is a left $R$-module, then $(\{\text{submodules of }M\},\subseteq)$ is a modular lattice. Taking $M\!=\!R$, we deduce that $(\{\text{ideals of }R\},\subseteq)$ is a modular ...
6
votes
1answer
208 views

Lattices are congruence-distributive

$\newcommand{\r}[1]{\mathrel{#1}}$ First, a few definitions. Given a lattice $L$, a congruence on $L$ is an equivalence relation $\theta$, compatible with the lattice operations, i.e. if ...